3.904 \(\int \frac{x \tan ^{-1}(a x)^{5/2}}{(c+a^2 c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=161 \[ -\frac{15 \sqrt{\frac{\pi }{2}} \sqrt{a^2 x^2+1} \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{\tan ^{-1}(a x)}\right )}{4 a^2 c \sqrt{a^2 c x^2+c}}-\frac{\tan ^{-1}(a x)^{5/2}}{a^2 c \sqrt{a^2 c x^2+c}}+\frac{5 x \tan ^{-1}(a x)^{3/2}}{2 a c \sqrt{a^2 c x^2+c}}+\frac{15 \sqrt{\tan ^{-1}(a x)}}{4 a^2 c \sqrt{a^2 c x^2+c}} \]

[Out]

(15*Sqrt[ArcTan[a*x]])/(4*a^2*c*Sqrt[c + a^2*c*x^2]) + (5*x*ArcTan[a*x]^(3/2))/(2*a*c*Sqrt[c + a^2*c*x^2]) - A
rcTan[a*x]^(5/2)/(a^2*c*Sqrt[c + a^2*c*x^2]) - (15*Sqrt[Pi/2]*Sqrt[1 + a^2*x^2]*FresnelC[Sqrt[2/Pi]*Sqrt[ArcTa
n[a*x]]])/(4*a^2*c*Sqrt[c + a^2*c*x^2])

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Rubi [A]  time = 0.22684, antiderivative size = 161, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {4930, 4898, 4905, 4904, 3304, 3352} \[ -\frac{15 \sqrt{\frac{\pi }{2}} \sqrt{a^2 x^2+1} \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{\tan ^{-1}(a x)}\right )}{4 a^2 c \sqrt{a^2 c x^2+c}}-\frac{\tan ^{-1}(a x)^{5/2}}{a^2 c \sqrt{a^2 c x^2+c}}+\frac{5 x \tan ^{-1}(a x)^{3/2}}{2 a c \sqrt{a^2 c x^2+c}}+\frac{15 \sqrt{\tan ^{-1}(a x)}}{4 a^2 c \sqrt{a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

Int[(x*ArcTan[a*x]^(5/2))/(c + a^2*c*x^2)^(3/2),x]

[Out]

(15*Sqrt[ArcTan[a*x]])/(4*a^2*c*Sqrt[c + a^2*c*x^2]) + (5*x*ArcTan[a*x]^(3/2))/(2*a*c*Sqrt[c + a^2*c*x^2]) - A
rcTan[a*x]^(5/2)/(a^2*c*Sqrt[c + a^2*c*x^2]) - (15*Sqrt[Pi/2]*Sqrt[1 + a^2*x^2]*FresnelC[Sqrt[2/Pi]*Sqrt[ArcTa
n[a*x]]])/(4*a^2*c*Sqrt[c + a^2*c*x^2])

Rule 4930

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Simp[((d + e*x^2)^
(q + 1)*(a + b*ArcTan[c*x])^p)/(2*e*(q + 1)), x] - Dist[(b*p)/(2*c*(q + 1)), Int[(d + e*x^2)^q*(a + b*ArcTan[c
*x])^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, q}, x] && EqQ[e, c^2*d] && GtQ[p, 0] && NeQ[q, -1]

Rule 4898

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_)/((d_) + (e_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(b*p*(a + b*ArcTan[
c*x])^(p - 1))/(c*d*Sqrt[d + e*x^2]), x] + (-Dist[b^2*p*(p - 1), Int[(a + b*ArcTan[c*x])^(p - 2)/(d + e*x^2)^(
3/2), x], x] + Simp[(x*(a + b*ArcTan[c*x])^p)/(d*Sqrt[d + e*x^2]), x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[e,
c^2*d] && GtQ[p, 1]

Rule 4905

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Dist[(d^(q + 1/2)*Sqrt[1
+ c^2*x^2])/Sqrt[d + e*x^2], Int[(1 + c^2*x^2)^q*(a + b*ArcTan[c*x])^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x
] && EqQ[e, c^2*d] && ILtQ[2*(q + 1), 0] &&  !(IntegerQ[q] || GtQ[d, 0])

Rule 4904

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Dist[d^q/c, Subst[Int[(a
+ b*x)^p/Cos[x]^(2*(q + 1)), x], x, ArcTan[c*x]], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[e, c^2*d] && ILtQ
[2*(q + 1), 0] && (IntegerQ[q] || GtQ[d, 0])

Rule 3304

Int[sin[Pi/2 + (e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[2/d, Subst[Int[Cos[(f*x^2)/d],
x], x, Sqrt[c + d*x]], x] /; FreeQ[{c, d, e, f}, x] && ComplexFreeQ[f] && EqQ[d*e - c*f, 0]

Rule 3352

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelC[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)])/
(f*Rt[d, 2]), x] /; FreeQ[{d, e, f}, x]

Rubi steps

\begin{align*} \int \frac{x \tan ^{-1}(a x)^{5/2}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx &=-\frac{\tan ^{-1}(a x)^{5/2}}{a^2 c \sqrt{c+a^2 c x^2}}+\frac{5 \int \frac{\tan ^{-1}(a x)^{3/2}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx}{2 a}\\ &=\frac{15 \sqrt{\tan ^{-1}(a x)}}{4 a^2 c \sqrt{c+a^2 c x^2}}+\frac{5 x \tan ^{-1}(a x)^{3/2}}{2 a c \sqrt{c+a^2 c x^2}}-\frac{\tan ^{-1}(a x)^{5/2}}{a^2 c \sqrt{c+a^2 c x^2}}-\frac{15 \int \frac{1}{\left (c+a^2 c x^2\right )^{3/2} \sqrt{\tan ^{-1}(a x)}} \, dx}{8 a}\\ &=\frac{15 \sqrt{\tan ^{-1}(a x)}}{4 a^2 c \sqrt{c+a^2 c x^2}}+\frac{5 x \tan ^{-1}(a x)^{3/2}}{2 a c \sqrt{c+a^2 c x^2}}-\frac{\tan ^{-1}(a x)^{5/2}}{a^2 c \sqrt{c+a^2 c x^2}}-\frac{\left (15 \sqrt{1+a^2 x^2}\right ) \int \frac{1}{\left (1+a^2 x^2\right )^{3/2} \sqrt{\tan ^{-1}(a x)}} \, dx}{8 a c \sqrt{c+a^2 c x^2}}\\ &=\frac{15 \sqrt{\tan ^{-1}(a x)}}{4 a^2 c \sqrt{c+a^2 c x^2}}+\frac{5 x \tan ^{-1}(a x)^{3/2}}{2 a c \sqrt{c+a^2 c x^2}}-\frac{\tan ^{-1}(a x)^{5/2}}{a^2 c \sqrt{c+a^2 c x^2}}-\frac{\left (15 \sqrt{1+a^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{\cos (x)}{\sqrt{x}} \, dx,x,\tan ^{-1}(a x)\right )}{8 a^2 c \sqrt{c+a^2 c x^2}}\\ &=\frac{15 \sqrt{\tan ^{-1}(a x)}}{4 a^2 c \sqrt{c+a^2 c x^2}}+\frac{5 x \tan ^{-1}(a x)^{3/2}}{2 a c \sqrt{c+a^2 c x^2}}-\frac{\tan ^{-1}(a x)^{5/2}}{a^2 c \sqrt{c+a^2 c x^2}}-\frac{\left (15 \sqrt{1+a^2 x^2}\right ) \operatorname{Subst}\left (\int \cos \left (x^2\right ) \, dx,x,\sqrt{\tan ^{-1}(a x)}\right )}{4 a^2 c \sqrt{c+a^2 c x^2}}\\ &=\frac{15 \sqrt{\tan ^{-1}(a x)}}{4 a^2 c \sqrt{c+a^2 c x^2}}+\frac{5 x \tan ^{-1}(a x)^{3/2}}{2 a c \sqrt{c+a^2 c x^2}}-\frac{\tan ^{-1}(a x)^{5/2}}{a^2 c \sqrt{c+a^2 c x^2}}-\frac{15 \sqrt{\frac{\pi }{2}} \sqrt{1+a^2 x^2} C\left (\sqrt{\frac{2}{\pi }} \sqrt{\tan ^{-1}(a x)}\right )}{4 a^2 c \sqrt{c+a^2 c x^2}}\\ \end{align*}

Mathematica [C]  time = 0.191129, size = 139, normalized size = 0.86 \[ \frac{15 i \sqrt{a^2 x^2+1} \sqrt{-i \tan ^{-1}(a x)} \text{Gamma}\left (\frac{1}{2},-i \tan ^{-1}(a x)\right )-15 i \sqrt{a^2 x^2+1} \sqrt{i \tan ^{-1}(a x)} \text{Gamma}\left (\frac{1}{2},i \tan ^{-1}(a x)\right )+4 \tan ^{-1}(a x) \left (-4 \tan ^{-1}(a x)^2+10 a x \tan ^{-1}(a x)+15\right )}{16 a^2 c \sqrt{a^2 c x^2+c} \sqrt{\tan ^{-1}(a x)}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(x*ArcTan[a*x]^(5/2))/(c + a^2*c*x^2)^(3/2),x]

[Out]

(4*ArcTan[a*x]*(15 + 10*a*x*ArcTan[a*x] - 4*ArcTan[a*x]^2) + (15*I)*Sqrt[1 + a^2*x^2]*Sqrt[(-I)*ArcTan[a*x]]*G
amma[1/2, (-I)*ArcTan[a*x]] - (15*I)*Sqrt[1 + a^2*x^2]*Sqrt[I*ArcTan[a*x]]*Gamma[1/2, I*ArcTan[a*x]])/(16*a^2*
c*Sqrt[c + a^2*c*x^2]*Sqrt[ArcTan[a*x]])

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Maple [F]  time = 0.946, size = 0, normalized size = 0. \begin{align*} \int{x \left ( \arctan \left ( ax \right ) \right ) ^{{\frac{5}{2}}} \left ({a}^{2}c{x}^{2}+c \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*arctan(a*x)^(5/2)/(a^2*c*x^2+c)^(3/2),x)

[Out]

int(x*arctan(a*x)^(5/2)/(a^2*c*x^2+c)^(3/2),x)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arctan(a*x)^(5/2)/(a^2*c*x^2+c)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arctan(a*x)^(5/2)/(a^2*c*x^2+c)^(3/2),x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*atan(a*x)**(5/2)/(a**2*c*x**2+c)**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x \arctan \left (a x\right )^{\frac{5}{2}}}{{\left (a^{2} c x^{2} + c\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arctan(a*x)^(5/2)/(a^2*c*x^2+c)^(3/2),x, algorithm="giac")

[Out]

integrate(x*arctan(a*x)^(5/2)/(a^2*c*x^2 + c)^(3/2), x)